... and beyond

A: magnitude of x coordinate of particle is 4m
B: magnitude of a average velocity is equal to average speed, for time interval t=0 to t=2sec
C: average acceleration is equal to instantaneous acceleration during interval t=0 to t=2sec
D: distance traveled in interval t=0 to t=4sec is 8m.
The question has multiple answers.

1 Answer

Answer:

Explanation:

I assume this refers to the position of the particle at #t=0#, since it travels along the x-axis only. We can substitute #0# into the given equation for position and see what we get for x.

#color(blue)(x(t)=(4t-t^2-4))#

#=>x(0)=(4(0)-(0)^2-4)#

#=>x=-4#

Therefore, the initial position of the particle is #-4"m"#, and #abs(-4)=4#, so answer A is correct.

#=>#B: The magnitude of the average velocity is equal to the average speed for the time interval #t=0# to #t=2# seconds.

We can compute the average velocity and average speed independently and compare the values we obtain.

The most important distinction between the two quantities is that the average speed is concerned with the distance the object travels over a given time period, whereas the average velocity is concerned with the displacement of the object over a given time period.

#color(blue)(v_"avg"=(Deltax)/(Deltat))#

We will first calculate the displacement #Deltax#, where #Deltax=x_f-x_i#.

#x_i=x(0)=-4"m"#

#x_f=x(2)=0"m"#

Therefore, #Deltax=0-(-4)=4"m"#.

We are given #Deltat=t_f-t_i=2-0=2"s"#

#v_"avg"=(4"m")/(2"s")#

#v_"avg"=2"m"//"s"# (in the direction of the positive x-axis)

Now we calculate the average speed:

#color(blue)(s_"avg"=(Deltad)/(Deltat))#

At #t=0#, the particle is at #x(0)=-4"m"#

At #t=1#, the particle is at #x(1)=-1"m"#

At #t=2#, the particle is at #x(2)=0"m"#

So, the object traveled a total distance of #(3+1)"m"=4"m"#

Therefore, #Deltad=4"m"# and the time period is still #2"s"#

#s_"avg"=(4"m")/(2"s")#

#s_"avg"=2"m"//"s"#

Therefore, #v_"avg"=s_"avg"# and answer B is correct.

#=>#C: The average acceleration is equal to the instantaneous acceleration during the time interval #t=0# to #t=2#.

We can find the instantaneous acceleration by taking the second derivative of the given equation for position. The average acceleration can be found as the change in velocity over time.

#color(blue)(a_"avg"=(Deltav)/(Deltat))#

We will first need to calculate #Deltav#, the change in velocity. We can find the velocity by taking the first derivative of the given equation for position and using the given time interval.

#v(t)=x'(t)=4-2t#

#v_i=v(0)=4#

#v_f=v(2)=0#

Therefore, #Deltav=0-4=-4"m"//"s"#.

#a_"avg"=(-4"m"//"s")/(2"s")#

#a_"avg"=-2"m"//"s"^2#

We can find the instantaneous acceleration by taking the derivative of the equation for velocity that we derived above, which is the second derivative of position.

#a(t)=v'(t)=-2"m"//"s"^2#

We see that #veca=a_"avg"# and therefore answer C is correct.

#=>#D: The distance traveled in the interval #t=0# to #t=4# seconds is #8"m"#.

We found above that the distance traveled by the particle when #tin[0,4]# was #4"m"#, so we know that at #x(2)#, the particle has already traveled #4"m"#.

At #t=2#, the particle is at #x(2)=0"m"#

At #t=3#, the particle is at #x(3)=-1"m"#

At #t=4#, the particle is at #x(4)=4"m"#

So the particle travels an additional #(1+3)=4"m"#. Therefore, the total distance is #4+4=8"m"#, and answer D is correct.